Hello, how can I do a (for loop) for this equation to find more than one value of the CF ? Specifically, I mean more than one value for (xm)

function [CF]= Coherence_Factor(xm,N);

CF = ((abs (sum(xm))).^2) / (N* sum(abs(xm.^2)))

end

%Function call window

xm= [-1:4],N=length(xm);

xm=[-5:-2],N=length(xm);

xm=[-2:1],N=length(xm)

[CF]= Coherence_Factor(xm,N)

Answer 1

[VBBV]

xm= {[-1:4], [-5:-2],[-2:1]};

xm = 1×3 cell array
    {[-1 0 1 2 3 4]}    {[-5 -4 -3 -2]}    {[-2 -1 0 1]}

for k = 1:numel(xm)
N=length(xm(k));
CF(k)= Coherence_Factor(xm(k),N);
end
CF

CF = 1×3

2.6129 3.6296 0.6667

function [CF]= Coherence_Factor(xm,N);
CF = ((abs (sum(cell2mat(xm)))).^2) ./ (N* sum(abs(cell2mat(xm).^2)));
end

Answer 2

[Voss]

xm = {-1:4, -5:-2, -2:1};

Nx = numel(xm);
CF = zeros(1,Nx);
for k = 1:Nx
    N = numel(xm{k});
    CF(k) = Coherence_Factor(xm{k},N);
end
CF

CF = 1×3

0.4355 0.9074 0.1667

function CF = Coherence_Factor(xm,N)
    CF = abs(sum(xm)).^2 / (N*sum(abs(xm.^2)));
end

Note that if the xm are always guaranteed to be real-valued (as opposed to complex-valued), then the abs() calls are unnecessary:

xm = {-1:4, -5:-2, -2:1};

Nx = numel(xm);
CF = zeros(1,Nx);
for k = 1:Nx
    N = numel(xm{k});
    CF(k) = Coherence_Factor(xm{k},N);
end
CF

CF = 1×3

0.4355 0.9074 0.1667

function CF = Coherence_Factor(xm,N)
    CF = sum(xm).^2 / (N*sum(xm.^2));
end

And if N is always supposed to be the number of elements in xm, you can define N inside the function Coherence_Factor and avoid having the second input argument:

xm = {-1:4, -5:-2, -2:1};

Nx = numel(xm);
CF = zeros(1,Nx);
for k = 1:Nx
    CF(k) = Coherence_Factor(xm{k});
end
CF

CF = 1×3

0.4355 0.9074 0.1667

function CF = Coherence_Factor(xm)
    N = numel(xm);
    CF = sum(xm)^2 / (N*sum(xm.^2));
end

[Walter Roberson]

I wonder if it is supposed to be sum(abs(xm).^2) or abs(sum(xm.^2)) ?

The current sum(xm) having priority has different results if values could be a mix of negative and positive